2 years ago

Find the matrix B of the linear transformation T(x)=Ax with respect to the basis B = (v1, v2,...,vm): A= [0 1; 2 3] v1= [1; 2] v2= [1; 1]. I can make the [x]_B span...how do you get that final matrix????? (basis change)

Okay sorry. If I have a vector
\[\vec{x} = \left(\begin{matrix} x_1 \\ x_2 \end{matrix}\right) \]
that I want to express as a vector y with respect to the basis vectors v1 and v2, then
\[ \vec{x} = [\vec{v_1} \space \vec{v_2}]\vec{y} = V\vec{y} \]
so,
\[\vec{y} = V^{-1}\vec{x} \]
So, the linear transformation
\[T(x) = A\vec{x} \]
can also be written as
\[T(y) = AV\vec{y} \]
But at the end of the day, this quantity
\[ \left( AV\vec{y}\right) \]
is a vector with respect to the basis V. To make it a vector with respect to x again, I must apply V inverse:
\[T(x) = V^{-1}AV\vec{y} = B\vec{y} \]
so in conclusion, long story short,
\[B = V^{-1}AV \]

awesome awesome! very cool..thanks sooooo much!! :) one final question (sorry to be annoying..) but so I found V(inv) to be [-2 -4; -2 -5] and then when I go through V(inv)*A*V, I am dealing with [-2 -4; -2 -5] *[0 1; 2 3]*[1 1; 2 1] but I keep ending up with [-10 -6; -12 -7]. I'm pretty sure it's not my matrix multiplication or inverted matrix V that's the problem..do you notice if I'm using a vector out of place somewhere?

OH MY GOODNESS!!!!! haha okay, so I was a little ahead of myself in my notes..my ridic V(inv) above was actually my INCORRECT V(inv)*A already..that's what happens when your work gets all jumbled up, I guess. But my inverted V was wrong anyway..I guess I should take a step back and work on that now hahaha. THANK YOU SO MUCH!!! (exam in 15 hours..angsting so hard right now!) you're awesome! =D